Why the dimension of the range of linear map is smaller than the dimension of $W$?

Question

I am working on the proof of a map to a smaller dimensional space is not injective.

Proof:

Let $T \in \mathcal{L}(V,W)$. Then $$\text {dim null } T = \text {dim } V - \text {dim range } T \geq \text {dim } V - \text {dim } W \gt 0$$

I don't understand why $\text {dim } V - \text {dim range } T \geq \text {dim } V - \text {dim } W$. Or why $\ \text {dim range } T \leq \text {dim } W$?

I guess it is because as range $T$ is coming from $V$, at most you can get the dimension of $V$. As dim $V$ is less than dim $W$, dim range $T$ is also less than dim $W$. I don't know if it is correct, and some better way to look at it.

Answer 1

By definition, $\text{range} (T)$ is a subspace of $W$. $$\text{range} (T) \le W$$ So, $$\text{dim} \,\text{range} (T) \le \text{dim} W$$

Answer 2

It's because $\operatorname{range}T$ is a subspace of $W$. Therefore, $\dim\operatorname{range}T\leqslant\dim W$.

Answer 3

Let me address what I think is giving you more of a problem thinking about this:

In vector spaces we have the notion of a basis; that is a set of vectors which are linearly independent and span the whole space. The smallest cardinality (size) of such a basis, which is invariant to the space, is called the dimension.

Now in your case, once you impose the image of $V$ in $W$ (and get a linear space in itself) and ask what the dimension of this image is, you're asking what is the minimal number of vectors I need to span the image. This is by definition less than the required number to span the space $W$ itself. Is this making the question more intuitive?

Answer 4

If $\text{rank}(T) > \dim{W}$, then you can find a linearly independent set in $W$ that has more elements than the dimension of $W$!